Networks 2
Part 3
Prepared by: Dr. Raed Qadi & Dr. Manar Qamhieh
The content of these slides is collected from various sources (books, websites and author pages), no originality is claimed
1
Course Outline
●
Wide Area Networks (WANs)
○
○
○
●
●
Framing
Virtual LANs (VLANs)
Ethernet/ Optical Fiber
Cisco routing and Access Lists
Exterior Gateway Protocols
CDMA / WiFi
Bluetooth and WiMAX
Cellular Networks and Mobile IP
802.15.14: Zigbee, PAN networks
Network Security
○
○
○
HTTP / PHP connection / Servlets
Routing Issues
○
○
●
SNMP protocol / MIB structure
Wireless and Mobile Networks
○
○
○
○
Frame Relay and ATM
Network Programming
○
●
●
LAN Design Issues
○
○
○
Network Management
○
WAN technologies
○
●
Overview
Packet/Circuit/Cell switched Networks
Data Link and Network Protocols
●
○
○
○
Security principles
Principles of cryptology
Message Integrity and Endpoint
authentication
Secure TCP connection (SSL)
Network layer security
Firewalls
2
Data Link Layer Issues: Framing
● Data-link layer provides service to network layer and it uses the service
provided by the physical layer
● The physical layer accepts a raw bit stream and attempts to deliver it to the
destination
○
○
○
The physical layer adds some redundancy to its signals to reduce the bit error rate
However, bit stream received by the data link layer is not error free (changed values, different
number of bits received)
Data link layer detects the errors and even corrects them
● Usually, the data link layer breaks up the bit stream into discrete frames,
computes a checksum for each frame and include it in the frame. The
checksum is recomputed at destination to detect errors.
3
Data Link Layer Issues: Framing
● Breaking up the bit stream into frames is not trivial.
○
○
It should be easy for a receiver to find the start of new frames
It should use little of bandwidth
● There are four methods for framing:
○
○
○
○
Byte count
Flag bytes with byte stuffing
Flag bits with bit stuffing
Physical layer coding violations ( Example: Manchester Encoding, Using more bits than
needed, using more levels than needed)
4
Framing: Byte Count
● Byte count method uses a field in the header to specify the number of bytes in
the frame.
● The data link layer at destination sees the byte count and knows how many
bytes follow and where the end of the frame is.
● This method is rarely used by itself
5
Framing: Byte Count
●
●
●
●
The trouble with this method is that the count can be modified due to errors.
Accordingly, the destination will get out of synchronization and it will be unable to locate
the correct start of the next frame
Even if checksum is incorrect and the destination knows that the frame is bad, it still
has no way of telling where the next frame starts
Ask for retransmission does not help since the destination does not know how many
bytes to skip over to get to start of retransmission.
6
Framing: Flag Bytes with Byte Stuffing
● This method have each frame start and end with flag byte which is used as
the starting and ending delimiter. In general, SOF and EOF can be different
bytes or same.
● Two consecutive flag bytes indicate the end of one frame and the start of the
next
● If the receiver ever loses synchronization, it can just search for two flag bytes
to find the end of the current frame and the start of the next one
7
Framing: Flag Bytes with Byte Stuffing
● Problem:
○
It may happen that the flag byte occurs in the data, which will interfere with the framing
● Solution: Byte stuffing
○
○
○
The sender’s data link layer can insert a special escape byte (ESC) just before each flag byte
in the data
Framing flag byte can be distinguished from one in the data by the absence or presence of an
escape byte before it
The data link layer on the receiver end removes the escape bytes before giving the data to the
network layer
● This method is a slight simplification of the approach used in PPP which is
used to carry packets over communication links
8
Framing: Flag Bytes with Byte Stuffing
9
Framing: Byte Stuffing General
● In the previous slides, we used SOF = and EOF were the same, and were both equl to
some byte, usually we use 7E = 01111110. But in general we can choose any 3 bytes
to be SOF, EOF and Escape
● Byte stuffing: Use the following rules
○
○
○
○
If SOF occurs in data, replace with Escape SOF
If EOF occurs in data, replace with Escape EOF
If Escape occurs in data, replace with Escape Escape
SOF, EOF are not preceded by Escape at the start and end of Frame, respectively
● Example:
● Use SOF = <, EOF = >, Escape = \
● show the transmitted frame if the data is : AB<<\CK\\z><\K>
● The transmitted Frame =
<AB\<\<\\CK\\\\z\>\<\\K\>>
● Notice that the added green < amd > are the SOF and the EOF respectively.
● When the frame is received, it will b destuffed and we restore original data which is AB<<\CK\\z><\K>
10
Framing: Flag Bytes with Bit Stuffing
● Framing is done at bit level (frames contain an arbitrary number of bits made
up of units of any size)
○
This gets around the disadvantage of byte stuffing which is that it is tied to the use of 8-bit
bytes
● It was developed for HDLC protocol
○
○
○
○
Each frame starts and ends with a flag byte of special pattern (01111110)
Whenever the sender’s data link layer encounters five consecutive 1s in the data, it stuffs a 0
bit.
This method ensures a minimum density of transitions
Whenever the receiver finds five 1s followed by a 0 bit, it deletes the 0
● USB (Universal Serial Bus) uses bit stuffing
11
Framing: Flag Bytes with Bit Stuffing
12
Byte and Bit Stuffing
● The main side-effect of byte and bit stuffing methods is that the length of a
frame depends now on the content of the data frame it carries
● A frame of 100 bytes
○
○
○
can be carried in a frame of 100 bytes if there are no flags in the data
can be carried in a frame of 200 bytes if the data consists solely of flag bytes and byte stuffing
is used
can be carried in a frame of a length increased by around 12.5% if the data consists solely of
flag bytes and bit stuffing is used
13
Framing: Physical Layer Coding Violations
● This method uses a shortcut from the physical layer
● The use of reserved signals to indicate the start and the end of frames
○
using code violations to delimit frames
● By using reserved signals, it is easy to find the start and end of frames and
there is no need to stuff data
● It is applicable to networks in which the encoding on the physical medium
contains some redundancy
○
In such cases normally, a 1 bit is a high-low pair and a 0 bit is a low-high pair. The combinations of low-low
and high-high which are not used for data may be used for marking frame boundaries.
14
Course Outline
●
Wide Area Networks (WANs)
○
○
○
●
●
Framing
Virtual LANs (VLANs)
Ethernet/ Fiber Optics
Cisco routing and Access Lists
Exterior Gateway Protocols
CDMA / WiFi
Bluetooth and WiMAX
Cellular Networks and Mobile IP
802.15.14: Zigbee, PAN networks
Network Security
○
○
○
HTTP / PHP connection / Servlets
Routing Issues
○
○
●
SNMP protocol / MIB structure
Wireless and Mobile Networks
○
○
○
○
Frame Relay and ATM
Network Programming
○
●
●
LAN Design Issues
○
○
○
Network Management
○
WAN technologies
○
●
Overview
Packet/Circuit/Cell switched Networks
Data Link and Network Protocols
●
○
○
○
Security principles
Principles of cryptology
Message Integrity and Endpoint
authentication
Secure TCP connection (SSL)
Network layer security
Firewalls
15
VLANs: Existing Shared LAN Configurations
● A Virtual LAN (VLAN) is a logical grouping of devices or users that can be
grouped by function, department or application, regardless of their physical
segment location.
● VLAN configuration is done at the switch via software.
● VLANs are not standardized and they require the use of proprietary software
from the switch vendor.
● VLANs logically segment the physical LAN infrastructure into different
subnets (or broadcast domains). Broadcast frames are switched only
between ports within the same VLAN.
16
VLANs
17
Typical LAN
● Typical LAN is configured according to the physical infrastructure it is
connecting.
○
○
○
○
Users are grouped based on their location in relation to the hub they are plugged in to and
how the cable is run to the wiring closet.
The router interconnecting each shared hub typically provides segmentation and can act as a
broadcast firewall (unlike segmentation provided by switches).
This typical segmentation does not group users according to their workgroup association or
their need for bandwidth.
Users share the same segment and the same bandwidth despite their workgroup or
department.
18
VLANs and Physical Boundaries
VLAN technology is a costeffective and efficient way of
grouping network users into
virtual groups regardless of their
physical location on the network.
Switch ports and connected
users can be grouped logically
into defined workgroups:
- co-workers in the same
department
- cross-functional product
team
- Diverse user groups
sharing the same network
application or software
19
LAN vs. VLAN Segmentation
● Main differences:
○
○
○
○
○
VLANs work at Layer 2 and Layer 3 of the OSI model
Communication between VLANs is provided by Layer 3 routing
VLANs provide a method of controlling network broadcasts
The network administrator assigns users to VLANs
VLANs can increase network security by defining which network nodes can communicate with
each other
20
LAN vs. VLAN Segmentation
Users are grouped on a
single switch or on connected
switches
VLANs can span singlebuilding infrastructures,
interconnected buildings or
even WANs
21
The transport in VLANs
● It is important to transport
VLAN information between
interconnected switches
and routers that reside on
the corporate backbone
● Transport capabilities:
○
○
○
remove physical boundaries
between users
increase the configuration flexibility of a VLAN solution when users move
provide mechanisms for interoperability between backbone system components
22
The transport in VLANs
● The backbone acts as
the collection point for
large volumes of traffic
● It carries end-user VLAN
information and
identification between
switches, routers and and
directly attached servers
● Within backbones, high-bandwidth, high-capacity links are typically chosen to
carry the traffic.
23
Routers in LAN
●
●
●
●
A router provides firewalls, broadcast management and route processing and
distribution
In VLAN, routers are used to provide connected routes between different VLANs.
They also connect VLAN to a part of the network that is logically segmented with
traditional subnet approach or require to access across a WAN
Layer 3 communication is either embedded in the switch or provided externally.
○
Integration of external routers is done by using one or more high-speed backbone connections
(like Fast Ethernet or ATM connections)
■ Increasing the throughput between switched and routers
■ Combine the overall number of physical router ports required for communication
between VLANs
24
Frames in VLAN
● A switch in VLAN communications filter and forward decisions between
switches and routers of the network by frame based VLAN metrics defined by
network managers
● The logical grouping of users into VLANs is done by frame filtering and
frame identification (frame tagging).
○
A frame is received or forwarded by a switch based on a set of rules defined by the
administrator
25
Frame Filtering
● Frame filtering examines particular information about each frame.
● A filtering table is developed for each switch
○
this provides a high level of administrative control because a switch can examine many
attributes of each frame
● A LAN switch can group users based on a station’s MAC address or protocol
type.
● The switch compares the frames it filters with table entries and it takes the
appropriate action based on the entries
○
This method is not scalable because each frame had to be referenced to a filtering table.
26
27
Frame Identification (tagging)
●
Frame Tagging uniquely assigns a VLAN ID to each frame.
○
●
VLAN IDs are assigned to each VLAN in the switch configuration by the switch
administrator
○
●
●
●
VLAN ID is a unique identifier added in the header of each frame as it is forwarded throughout
the network backbone
this method is a more scalable solution to VLAN deployment than frame filtering
The ID is understood and examined by each switch prior to any broadcasts or
transmissions to other switches, routers or end-station devices
When the frame exits the network backbone, the switch removes the identifier before
the frame is transmitted to the target end station
Frame tagging functions at Layer 2 and requires little processing or administrative
overheads.
28
29
VLAN Ports and Broadcast
● In a VLAN switch, each port can be assigned to a VLAN. Ports assigned to
the same VLAN share broadcasts.
○
This improves the overall performance of the network
● There are three VLAN implementation methods to assign a switch port to a
VLAN:
○
○
○
Port-centric
Static
Dynamic
30
PortCentric
VLAN
All nodes connected to ports in the same
VLAN are assigned to the same VLAN ID
Port-Centric facilitates the
administrator's job:
-
Users are assigned by
port
VLANs are easily
administered
Increased security
between VLANs
Packets don’t leak on
other domains
31
Static VLAN
Advantages of static VLAns:
-
They are secure
easy to configure
straightforward to monitor
Disadvantages:
-
They require the control
and management of an
administrator
Ports on a switch are statically assigned to a VLAN. These
ports maintain their assigned configurations until changed by
the administrator
32
Dynamic VLAN
VLANs are assigned based on a
centralized VLAN management
application.
When a station is connected to
an unassigned port, the switch
checks the MAC address entry
in the VLAN database.
The switch dynamically
configures the port with the
corresponding VLAN
configuration
Switch ports can automatically determine their VLAN
assignments. Dynamic VLAN functions are based on MAC
addresses, logical addressing or protocol type of data
33
Dynamic VLAN
Advantages:
-
-
less administration within the
wiring when a new user is
added or moved
centralized notification when
unrecognized user is added
Disadvantages:
-
More administration is
required to set up database
and to maintain an accurate
database of all users
Switch ports can automatically determine their VLAN
assignments. Dynamic VLAN functions are based on MAC
addresses, logical addressing or protocol type of data
34
Benefits of VLANs
● VLANs make user additions, moves and changes easier
○
○
These changes are one of the network manager’s biggest headaches and largest expenses
Many moves require cabling, and almost all moves require new station addressing and hub
and router reconfigurations
● VLANs provide an effective mechanism for controlling these changes and
reducing cost associated with hub and router reconfigurations
○
○
○
Users in the same VLAN share the same network address space regardless of their location
When users in a VLAN are moved from one location to another (without changing VLAN and
they are already connected to a switch port), their network addresses do not change.
A location change can be as simple as plugging a user into a port on a VLAN-capable switch
and configuring the port on the switch to that VLAN.
35
Broadcast in VLAN
●
●
Broadcast frequency depends on types of
applications, types of servers, the amount
of logical segmentation and how network
resources are used.
Firewall segmentation is an effective
measure against the problems of
broadcast.
○
○
Firewalls segment the network to prevent
one segment from damaging other parts
of the network
It provides reliability and minimizes
overheads of broadcast traffic
Broadcasts (Layer 2 transmissions) can result
from multimedia applications of faulty devices.
They can bring down network.
Broadcasts need boundaries
36
●
VLANs are effective
mechanism for extending
firewalls from the routers to
switches in flat networks
○
●
A flat network has one
broadcast domain across the
entire network (no routers
between switches)
VLANs create firewalls by
assigning switch ports or users
to specific VLAN groups (within
single switches or across
multiple connected switches)
○
Broadcast traffic within one
VLAN is not transmitted
outside the VLAN
37
IEEE 802.1Q Standard
The IEEE 802.1Q Standard was published in 1998 and it defines a new format to contain a
VLAN tag. This format can be used with old format 802.3 in the same network based on
available equipments
38
IEEE 802.1Q Standard
IEEE 802.1Q frames are understood by VLAN-aware switches and the VLAN fields are filled by the first
VLAN-aware switch to touch the frame.
Legacy switches are moot regarding 802.1Q frames since such frames are not supposed to be sent to
these switches.
39
IEEE 802.1Q Standard
VLAN protocol ID has always the value 0x8100. All Ethernet cards recognize this
field as a type rather than length (since this number is greater than 1500)
40
IEEE 802.1Q Standard
The Tag is a 2-byte field that contains 3 subfields.
41
IEEE 802.1Q Standard
Priority field (3 bits) is unrelated to VLANs. It is used to distinguish hard real-time traffic from
soft real-time traffic from time-insensitive traffic in order to provide better quality of service
over Ethernet.
42
IEEE 802.1Q Standard
CFI (Canonical Format Indicator) was originally intended to indicate the order of bits in the
MAC address. Now it is used to indicate that the payload contains an 802.5 frame that is
hoping to find another 802.5 frame at the destination while being carried by Ethernet.
43
IEEE 802.1Q Standard
VLAN identifier (low-order 12 bits) identifies the VLAN to which the frame belongs
44
VLANs and Network Security
● LANs are widely used and they have confidential, mission-critical data moving
across them
● Confidential data requires security through access restriction
● LANs are easy to penetrate
○
A user plugged in to a live port can have access to all traffic within the segment.
● Network segmentation into multiple broadcast groups is a solution:
○
○
○
Restricts the number of users in a VLAN group
Prevent another user from joining without first receiving approval from the VLAN network
management application
Configure all unused ports to a default low-service VLAN
45
VLANs and
Network Security
Restricted applications
resources are grouped in a
secured VLAN (a switch
restricts access into the group
based on station addresses,
application types or protocol
types)
More security can be added by
using access control lists
46
Course Outline
●
Wide Area Networks (WANs)
○
○
○
●
●
Framing
Virtual LANs (VLANs)
Ethernet/ Optical Fiber
Cisco routing and Access Lists
Exterior Gateway Protocols
CDMA / WiFi
Bluetooth and WiMAX
Cellular Networks and Mobile IP
802.15.14: Zigbee, PAN networks
Network Security
○
○
○
HTTP / PHP connection / Servlets
Routing Issues
○
○
●
SNMP protocol / MIB structure
Wireless and Mobile Networks
○
○
○
○
Frame Relay and ATM
Network Programming
○
●
●
LAN Design Issues
○
○
○
Network Management
○
WAN technologies
○
●
Overview
Packet/Circuit/Cell switched Networks
Data Link and Network Protocols
●
○
○
○
Security principles
Principles of cryptology
Message Integrity and Endpoint
authentication
Secure TCP connection (SSL)
Network layer security
Firewalls
47
LAN Issues: Broadcast Channels
● Network links can be divided into two categories:
○
○
Networks using Point-to-Point connections (as in WANs excepts for satellite networks)
Networks using broadcast channels (as in LANs and particularly wireless networks)
● Broadcast channels are referred to as multi-access channels or random
access channels
○
The broadcast channel connects each user to all other users, and any user who makes full
use of the channel interferes with other users who also wish to use the channel
● The MAC (Medium Access Control) protocols are used to solve the
interference problem of broadcast channels.
○
MAC sublayer is the bottom part of the data-link layer.
48
Channel Allocation Problem
● This is the problem of allocating a single broadcast channel among competing
users.
○
A channel can be a portion of wireless spectrum or a single wire or an optical fiber
● There are two methods of allocation:
○
○
Static allocation
Dynamic allocation
49
Static Channel Allocation
● The traditional way of static allocation of a single channel is the use of
multiplexing schemes.
○
Divide the capacity of the channel among multiple competing users
● An example of multiplexing schemes is the Frequency Division
Multiplexing (FDM)
○
○
The bandwidth is divided into N equal-size portions (where N is the number of users).
Each user has a private frequency band and there is no interference among users.
● When FDM is an efficient allocation mechanism?
○
When the number of users is small and constant and each of them has a steady stream or a
heavy load of traffic
■ Example: FM radio stations
50
Static Channel Allocation
● Limitations of static allocation mechanism:
○
○
When the number of senders is large and varying
■ If the spectrum is cut into N portions, some bandwidth will be wasted due to the fewer
number of users who are currently interested in communication
■ If more than N users want to communicate, some of them will be denied access for lack
of bandwidth even if there are some inactive users.
the traffic is bursty (short and separate bursts of traffic)
■ An inactive user holds his portion of bandwidth and no one else is allowed to use it
■ Static allocation is inefficient because computer systems have usually bursty data traffic
and usually most of the channels are idle most of the time.
51
Dynamic Channel Allocation
● Numerous dynamic channel allocation algorithms have been revised.
● ALOHA protocol (with or without slotting) is used in many derivatives in real
systems (cable modems and RFID)
● Carrier Sensing is a technique that can be used when the state of the channel
can be sensed. Using this technique, stations can avoid starting a
transmission while another station is transmitting.
○
○
This technique has led to a variety of CSMA protocols for LANs and MANs.
It is the basis for classical Ethernet and 802.11 (wireless) networks.
52
IEEE 802.3 (Ethernet)
● Ethernet is the most appearing kind of computer networks in the world
● There are two kinds of Ethernet exist:
○
○
Classic Ethernet
■ it is the original form of Ethernet and ran at rates from 3 to 10 Mbps
■ it solves the multiple access problem, it uses CSMA/CD
Switched Ethernet (Still CSMA/CD)
■ switches are used to connect different computers
■ the current forms of Ethernet that are used nowadays
■ it runs at rates 100, 1000 and 10,000 Mbps
■ it runs in forms called fast Ethernet, gigabit Ethernet and 10 gigabit Ethernet
53
Classic Ethernet: Physical Layer
Architecture of classical Ethernet(Obselete)
There were two versions of classical Ethernet: thick and thin Ethernet. Each version has a
maximum cable length per segment (500m for this, 200 for thin) and can handle a maximum
number of machines. Usually up to 4 repeaters so thick can be up to 2000m.
54
Classic Ethernet: Physical Layer
● Each version of Ethernet has a maximum cable length per segment (i.e.
unamplified length) over which the signal will propagate.
● Multiple cables in larger networks can be connected by repeaters.
○
A repeater is a physical layer device that receives, amplifies and retransmits signals in both
directions
● The network is a series of cable segments connected by repeaters, over
which the information was sent using the Manchester encoding.
○
○
No two transceivers could be more than 2.5km apart and no path between any two
transceivers could traverse more than four repeaters.
These restrictions are done so that the MAC protocol would work correctly.
55
Classic Ethernet: MAC Sublayer Protocol
●
●
Preamble contains the bit pattern 10101010 (with the exception of the last byte in
which the last 2 bits are set to 11 to tell the receiver that the rest of the frame is about
to start). In 802.3, the last byte is called Start Of Frame (SOF)
The manchester encoding of this pattern produces 10-MHz square wave for 6.4µs to
allow the receiver’s clock to synchronize with the sender.
Ethernet
IEEE 802.3
56
Classic Ethernet: MAC Sublayer Protocol
●
●
The first transmitted bit of the Destination address (6 bytes) is a 0 for ordinary
address and 1 for group address (multicasting)
- group addresses allow multiple stations in a group to listen to a single address
- it requires group management to define which stations are in the group
Special address of all 1 bits is reserved for broadcasting (FF FF FF FF FF FF).
Ethernet
IEEE 802.3
57
Classic Ethernet: MAC Sublayer Protocol
●
Source addresses (6 bytes) are globally unique addresses assigned by IEEE
- the first 3 bytes of the address are used for an OUI (Organizationally Unique
Identifier). Values are assigned by IEEE and indicate a manufacturer
- the last 3 bytes are assigned by the manufacturer who programs the complete
address into the NIC before it is sold.
Ethernet
IEEE 802.3
58
Classic Ethernet: MAC Sublayer Protocol
●
●
Type field in Ethernet frame tells the receiver what to do with the frame (which process
to give the frame to)
- Example: type code 0x0800 means that the data contains an IPv4 packet
In 802.3, this field indicates the length of the frame so as to prevent a layering violation
- Since Ethernet length was determined by looking inside data
Ethernet
IEEE 802.3
59
Classic Ethernet: MAC Sublayer Protocol
●
Both fields are accepted by IEEE.
- All type fields in use have values greater than 1500 (maximum length of data)
- Any number less than or equal to 1500 (0x600) can be interpreted as Length
- Any number greater than 1500 (0x600) can be interpreted as Type
- Checksum is CRC
Ethernet
IEEE 802.3
60
Classic Ethernet: MAC Sublayer Protocol
●
●
Data field is up to 1500 bytes (maximum frame length).
A minimum frame length is set to 46 bytes of data (a total of 64 bytes from destination
address to checksum including both)
-
It is necessary to distinguish valid frames from garbage (due to collisions)
To prevent a station from completing the transmission of a short frame before the first bit has
even reached the far end of the cable where it may collide with another frame
Ethernet
IEEE 802.3
61
Classic Ethernet: Minimum Frame Size
Collision detection can take as long as 2𝝉, which requires a minimum frame length to ensure that the sender is still
transmitting in order to detect the collison. The approcimate derivation above shows that Minimum Frame Size, Bit Rate,
and Maximum distance between Computers are related.
62
Ethernet: Min Frame size
●
●
●
●
●
●
From the derivation in the previous slide =>Min Frame size = K * d * Bit-rate
In10 Mbit Ethernet, Bit-Rate = 10 Mb/s, d = 2000 m( or 2500m).
Repeaters delay should be added, we ignored them in the derivation
The Min Frame Size = 64 bytes including header => Min Data = 64 bytes, hence we
use padding if data < 64 bytes.
In 100Mbit Ethernet Bit-Rate = 100 Mb/s, the distance was reduced to 200m hence
MinFrame Size Remains 64 bytes. If we keep it 2000m then MinFrame Size should be
640bytes. But the attenustion increases with Frequency and 2000m is not practical.
In Giga bit Ethernet , Bit-Rate = 1000Mb/s, if we keep the Maximum 200m, distance =>
Min Frame size = 640 bytes or we use Max distance = 20m and keep min Frame size =
64 byte. But 20m is not practical. So they decided to keep the max distance to 200m
and increase the Minimum frame size to around 600 byte.
63
Ethernet: Min Frame size
● So In 1Giga bit Ethernet Max distance = 200 m, Min Frame size should around 64*10
● 1Giga bit: In Fiber Networks they use Min Frame Size = 416 byte and 520 in Wired
Netwoeks
Network
10 Mbit
100 Mbit
1000 Mbit Giga
Max Distance between 2
Computers (Twisted pair)
2000m (old coax)
200 m new cat3
200 m cat 5
200 m Cat 5e
More in Fiber
Max Distance to Switch
(Twisted Pair)
100 m
100 m
100 m Cate 5e
Min Frame Size
64 byte
64 byte
520 byte for 1000 base T
416 bye In 1000 base X
64
Classic Ethernet: MAC Sublayer Protocol
●
Pad field is used to fill out the frame to the minimum size
-
This happens if the data portion of a frame is less than 64 bytes
Ethernet
IEEE 802.3
65
Classic Ethernet: MAC Sublayer Protocol
●
Checksum fields is a 32-bit CRC
-
It is an error-detecting code that is used to determine if the bits of the frame have been
received correctly
If an error is detected in a frame, the frame is dropped
Ethernet
IEEE 802.3
66
Classic Ethernet: CSMA/CD
● Classic Ethernet uses 1-persistent CSMA/CD
○
○
○
○
A transmitter senses the medium when it has frames to send.
It sends data as soon as the medium becomes idle
If there is a collision, transmitters abort transmission with a short jam signal and retransmits
after a random interval
If there is no collision, the sender assumes that the frame was probably successfully delivered.
Neither CSMA/CD nor Ethernet provides acknowledgments
■ This choice is appropriate for wired and optical fiber channels that have low error rates.
Wireless channels have more errors and hence, acknowledgments are used
67
Switched Ethernet
● Switched Ethernet has wiring pattern in which each station has a dedicated
cable running to a central hub (Obselete)
○
○
Instead of the single long cable architecture of classic Ethernet.
Switched Ethernet architecture solved the problems of classic Ethernet architecture, such as
finding breaks or loose connections
● Hubs do not increase capacity and Ethernet
network can become saturated
○
A hub is equivalent to the single cable of classic
Ethernet (equivalet to Coaxial cable = Bus)
68
Switched Ethernet
● A switch deals with the problem of
increased load.
● It contains a high-speed backplane that
connects all of the ports
● Unlike hubs, a switch only outputs frames to
the ports for which those frames are destined
○
○
○
When a switch port receives an Ethernet frame from
a station, the switch checks the Ethernet addresses
to see which port the frame is
destined for
The switch forwards the frame over its high-speed backplane to the destination port
Other ports on the switch do not know about this frame
● All configurations use point-to-point links.
69
Switched Ethernet: Collisions
● If more than one station send data at the same time:
○
○
In a hub, all stations are in the same collision domain
■ They must use the CSMA/CD algorithm to schedule their transmissions
In a switch, each port is its own independent collision domain
■ If the cable is full-duplex, both station and port can send data at the same time without
worrying about other stations or ports. No CSMA/CD is needed
■ If the cable is half-duplex, the station and the port must contend for transmission using
CSMA/CD
Full-Duplex Mode
70
Switched Ethernet: Hub vs Switch
● A switch improves performance over a hub in two ways:
○
○
There are no collisions when a switch is used unless it is broadcst or two try to send to the
same destination. Then the capacity is used more efficiently
Multiple frames can be sent simultaneously (by different stations)
■ These frames will reach the switch ports and travel over the switch’s backplane to be
output on the proper ports.
■ Switch has buffering so that it can temporarily queue an input frame until it can be
transmitted to the output port (if two frames are sent to the same output port)
● Network is more secure when switches are used rather than hubs
○
○
With a hub, every computer that is attached can see the traffic sent between all of the other
computers
With a switch, traffic is forwarded only to the ports where it is destined
71
Fast Ethernet (IEEE 802.3u)
● Fast Ethernet is developed to get faster LANs
● It is backward compatible with Ethernet
● Basic information:
○
○
keep all old frame formats, interfaces and procedural rules of Ethernet
Reduce the bit time from 100ns (1/10Mbps) to 10ns (100Mbps)
72
Fast Ethernet (IEEE 802.3u)
●
●
A Category 3 twisted pairs can carry 10 Mbps with 100m, but it is unable to carry 100
Mbps within the same distance
100Base-T4 is a Category 3 UTP scheme
○
●
It used a signaling speed of 25 MHz, which is 25% faster than standard Ethernet (Manchester
Encoding requires two clock periods for each sent bit)
100Base-T4 requires 4 twisted pairs. Half Duplex
○ One is always to the hub
○ One is always from the hub
○ Two are switchable to the current transmission direction
73
Fast Ethernet (IEEE 802.3u)
●
100Base-TX Ethernet is used based on Category 5 UTP
○ It dominated the market for Fast Ethernet since it can 100 Mbps transmission over
a 100m easily with a simple design
○ Two twisted pairs per station are used and they can handle clock rates of 125
MHz
■
○
It uses 4B/5B encoding instead of Manchester encoding
■
○
One to the hub and one from it
4 data bits are encoded as 5 signal bits and sent at 125 Mbit to provide 100 Mbps
The 100Base-TX is full duplex
■
Stations can transmit at 100 Mbps on one twisted pair and receive at 100 Mbps on
another twisted pair at the same time
74
Fast Ethernet (IEEE 802.3u)
● 100Base-FX uses two strands of multimode fiber (one for each direction)
● It can run full duplex with 100 Mbps in each direction
● The distance between a station and the switch can be up to 2 km
75
Physical Limitations of Fast Ethernet
● Since Fast Ethernet allows interconnection by either hubs or switches,
CSMA/CD continues to work and the relation between the minimum frame
size and maximum cable length must be maintained when speed goes from
10 Mbps to 100 Mbps
○
○
Either the minimum frame size of 64 bytes must go up
Or the maximum cable length of 2500m must come down
● The easiest solution (and the applied one) is to reduce the maximum cable
length between any two stations by a factor of 10
76
Fast Ethernet and Backward Compatibility
● Fast Ethernet was quickly deployed by many users, but some users kept their
old 10 Mbps Ethernet cards installed.
● Fast Ethernet can handle virtually a mix of 10Mbps and 100Mbps stations.
● Auto-Negotiation is a mechanism that lets two stations automatically
negotiate the optimum speed (10 or 100 Mbps) and duplexity (half or full)
● It works well most of the time but it is known to lead to duplex mismatch
problems when one end of the link auto-negotiates but the other one does not
and is set to full-duplex mode
● Most Ethernet products use this feature to configure themselves
77
Gigabit Ethernet (IEEE 802.3ab)
●
●
Gigabit Ethernet is a faster Ethernet than Fast Ethernet
All configurations used point-to-point links (each Ethernet cable has exactly two
devices on it).
Two-station Ethernet
Multistation Ethernet
78
Gigabit Ethernet: Duplexing
● Like Fast Ethernet, Gigabit Ethernet supports two different modes of
operation:
○
Full-duplex mode (normal mode) which allows traffic in both directions at the same time
■ It is used when a central switch is used.
■ All lines are buffered so each computer and switch is free to send frames whenever it
wants to
■ The sender does not have to sense the channel before transmission since contentions
are impossible
■ The CSMA/CD protocol is not necessary to be used
■ The maximum length of the cable is limited by the signal strength issues rather than by
how long it takes for a noise burst to propagate back to the sender in the worst case
■ Auto-negotiation is available (choices are 10, 100 and 1000 Mbps)
79
Gigabit Ethernet: Duplexing
○
Half-duplex mode is used when computers are connected to a hub rather than a switch
■ Hubs do not buffer incoming frames but it connects electrically all the lines internally.
■ When hubs are used, collisions are possible so the CSMA/CD protocol is necessary
● When CSMA/CD is used on wire CX gigabit Ethernet, the maximum length
should drop to 25m instead of 2500m in classic Ethernet.
● Two features are added to increase the maximum cable length to 200m:
○
○
Carrier extension: the hardware adds its own padding after the normal frame to extend the
frame to 512 bytes (instead of 64 bytes). The padding is added and removed by the hardware
without any changes done at software
Frame bursting: allows the sender to transmit a concatenated sequence of multiple frames in
a single transmission. If the total burst is less than 512 bytes, the hardware pads it again.
80
Gigabit Ethernet Cabling
●
●
gigabit Ethernet supports both copper and fiber cabling
Signaling at or near 1 Gbps requires encoding and sending a bit every nanosecond
○
The 8B/10B encoding is used, which encodes 8 bits of data into 10-bit codewords that are sent
over the wire or fiber. The codewords are chosen so that they could be balanced (have the
same number of 0s and 1s) with sufficient transitions for clock recovery
81
MAC Frame with Gigabit Carrier Extension
82
Minimu Frame Size, Distance, and Wiring
520 bytes applies to 1000Base-T implementations. The minimum frame size with extension field for
1000Base-X is reduced to 416 bytes because 1000Base-X encodes and transmits 10 bits for each byte
83
Gigabit Frame-Burst Sequence
Burst mode is a feature that allows a MAC to send a short sequence (a burst) of frames equal to
approximately 5.4 maximum-length frames without having to give up control of the medium. The
transmitting MAC fills each interframe interval with extension bits, so that other stations on the network will
see that the network is busy and will not attempt transmission until after the burst is complete.
84
Gigabit Frame-Burst Sequence
If the length of the first frame is less than the minimum frame length, an extension field is added to
extend the frame length to the minimum value.
Subsequent frames do not need extension fields, and a frame burst may continue as long as the burst
limit has not been reached. If the burst limit is reached after a frame transmission has begun, transmission
is allowed to continue until that entire frame has been sent.
85
Data Transmission
● Physical limits of a transmission system:
○
○
Propagation delay: time required for a segment to travel across media
bandwidth: maximum times per second a signal can change
● Network hardware encodes information for transmission
● There are two types of encoding:
○
○
Analog: the value of energy proportional to the amount of data sent
Digital: two forms of energy to encode (0 and 1)
■ Digital encoding is the one used in computer networks
86
Digital Encoding: Example
● Medium:
○
Copper wire
● Energy Form:
○
Electrical current
● Encoding:
○
○
Negative voltage encodes 1
Positive voltage encodes 0
87
Digital Encoding: RS-232C Standard
● Example uses:
○
○
Connection to keyboard/mouse
Serial port on PC
● Voltage is +15 or -15
● Cable limited to around 50 feet (around 15 m)
● Uses Asynchronous communication
88
Asynchronous Communication
● Sender and receiver must agree on
○
○
Number of bits per character
Duration of each bit
● Receiver
○
○
Does not know when a character will arrive
May wait forever
● To ensure meaningful exchange send
○
○
Start bit before character
One or more stop bits after character
89
Illustration of RS-232
● Start bit: same as 0 (not part of data)
● Stop bit: same as 1 (Follows data)
90
Duration of Bit in RS-232C
● Duration of bit is determined by baud rate
○
○
Typical baud rates: 9.6 Kbaud, 14.4 Kbaud and 28.8 Kbaud
Duration of 1 bit is 1/ baud-rate
● Sender and receiver must agree a priori
● Receiver samples signal
● Disagreement results in framing error
91
RS-232C: Two Way Communication
●
●
●
●
Communication is full-duplex, which is desirable in practice
Transmitter on one side connected to receiver on other
Separate wires needed to carry current in each direction
DB-25 connector used:
○
○
Pin 2 is transmit
Pin 3 is receive
92
Distorted Signal for a Single Bit
● RS-232 hardware must handle minor
distortions
○
○
Take multiple samples per bit
Tolerate less than full voltage
● Cannot use electrical current for long-distance
transmission
○
An oscillating signal travels farther than direct current
● For long-distance communication:
○
○
Send a sine wave (carrier wave)
Change (modulate) the carrier to encode data
93
Illustration of a Carrier
● Carrier:
○
Usually a sine wave and it oscillates continuously
● Fixed frequency of carrier
94
Types of Modulation
● Amplitude modulation (used in AM radio)
● Frequency modulation (used in FM radio)
● Phase-Shift modulation (used for data)
95
Amplitude Modulation
● Strength of signal encodes
0 or 1
● One cycle of wave needed
for each bit
● Data rate limited by carrier
bandwidth
96
Phase-Shift Modulation
● Change in phase encodes K bits
● Data rate higher than carrier bandwidth
97
Phase-Shift Example
● Section of wave is omitted at
phase shift
● Data bits determine size of
omitted section
98
Modem
● Hardware device
● Used for long-distance communication
● Contains separate circuitry for
○
○
Modulation of outgoing signal
Demodulation of incoming signal
● Modem stands for Modulator/Demodulator
99
Modems Used Over a Long Distance
● One modem at each end
● Separate wires carry signals in each direction
● Modulator on one modem connects to demodulator on other
100
Types of Modems
● Conventional
○
Use four wires and transmit modulated electrical wave
● Optical
○
Use glass fiber and transmit modulated light
● Wireless
○
Use air/space and transmit modulated RF wave
● Dialup
○
Use voice telephone system and transmit modulated audio tone
101
Dial-Up Modem
● Modem can
○
○
Dial
Answer
● Carrier is audio tone
102
Maximum Data Rate of a Channel
● The nature of the physical layer imposes two fundamental limits on all
channels, and these determine their bandwidth:
○
○
The Nyquist limit, which deals with noiseless channels
■ Henry Nyquist realized that even a perfect channel has a finite transmission capacity
The Shannon limit, which deals with noisy channels
■ Claude Shannon carried Nyquist’s work further and extended it to the case of a channel
subject to random noise
103
Nyquist’s Theorem
● Nyquist proved that if an arbitrary signal has been run through a low-pass
filter of bandwidth B, the filtered signal can be completely reconstructed by
making only 2B (exact) samples per second.
○
Sampling the line faster than 2B times is pointless, since the higher-frequency components
that such sampling could recover have already been filtered out
● If the signal consists of V discrete levels, Nyquist’s theorem states:
maximum data rate = 2B log2V bits/sec
104
Nyquist’s Theorem: Applications
● For RS-232
○
○
The number of discrete levels V is only 2, because RS-232 uses two values only (+15,-15
volts) to encode data bits
Data rate = 2 * B * log22 = 2B
● For Phase-Shift encoding
○
○
Suppose it has 8 discrete levels (possible shifts)
Data rate = 2 * B * log28 = 6B
105
Nyquist’s Theorem
● Example:
○
A noiseless 3-KHz channel of two levels have a maximum data rate of:
maximum data rate = 2B log2V bits/sec
maximum data rate = 2 * 3,000 * log22
= 6,000 bps
106
Shannon’s Theorem
● Shannon’s theorem considers the presence of random noise in channels
● According to physics, real systems emit and absorb energy (thermal).
Unwanted energy is called noise
○
Random (thermal) noise is due to the motion of the molecules in the system
● The amount of thermal noise present is measured by the ratio of the signal
power to the noise power, called SNR (Signal-to-Noise Ratio)
○
Usually the ratio is expressed on a log scale
107
Shannon’s Theorem
● According to Shannon’s theorem, the maximum data rate (or capacity) of a
noisy channel whose bandwidth is B Hz and whose signal-to-noise ratio is
S/N, is given by:
maximum number of bits/sec = B log2 (1+S/N)
***
This data rate describes the best capacities that real channels can have.
108
Shannon’s Theorem: Modem Example
●
Conventional telephone system:
○
○
○
○
Engineered for voice
Bandwidth (B) is 3000 Hz
Signal-to-Noise Ratio (SNR) is approximately 1000
Effective capacity is
Data rate = 3000 * log2(1+1000) ≅ 30,000 bps
● Conclusion
○
Dial-up modems have little hope of exceeding 28.8 Kbps , for B = 3K. For countries that limit
B=4K Hz. Maximum Data Rate = 40 Kbps
109
Shannon’s Theorem: ADSL Example
●
ADSL provides internet access over normal telephone lines.
○
○
○
○
It uses a bandwidth of around 1 MHz.
The SNR depends strongly on the distance of the home from the telephone exchange
(an SNR of around 40 dB for short lines of 1 to 2 km is very good)
With these characteristics, the channel can never transmit much more than 13 Mbps, no
matter how many or how few signal levels are used, and no matter how often or how
infrequently samples are taken.
For ADSL to exceed 13 Mbps, it must either:
■ improve the SNR (for example by inserting digital repeaters in the lines closer to the
customers)
■ or use more bandwidth (like the evolution to ADSL2+)
110
Conclusion: Nyquist vs Shannon
● Nyquist’s theorem means that finding a way to encode more bits per cycle
improves the data rate
● Shannon’s theorem means that no amount of clever engineering can
overcome the fundamental physical limits of a real transmission system
111
Nyquist and Shannon Example
● Consider a communication channel with Band width 5MHz and S/N = 511
(around 500), answer the following questions:
1. Compute the Capacity of the Channel which is the Maximum Possible Bit Rate
2. What is the Bit Rate if we use 8 levels of Encoding.
3. What is the Maximum Possiblle Level K such that n bits can be encoded where 2n = K. Where n is Integer.
● Solution :
1. Use Shannon. Capacity = maximum bit Rate bits/sec = B log2 (1+S/N) = 5M log2 (1+511) = 5*9 = 45M bit/sec
2. use Nyquist and Make sure Bit Rate is less than Capacity What is the Bit Rate if we use 8 levels of Encoding.
3.
Bit Rate = 2B log2 (Levels) = 2*5 log2 (8) = 30 Mbps which is less than 45Mbps(baud rate = 2B= 10M baud)
2B log2 (Levels) = B log2 (1+S/N) => Levels = sqrt(1+S/N) = 22.62 , since we want power of 2 ➔ Levels =
16 ➔ n= 4. Use 16 levels where each encode 4 bits. It is possible to use More levels up to 22 and these
levels can be used for error detection, correction and encoding of SOF, EOF if desired.
112
Ethernet Physical Layer
● The naming convention of any Ethernet product is a concatenation of three
terms indicating:
○
○
○
The transmission rate
The transmission method
The media type/signal encoding
● For example:
○
○
○
○
10Base-T
100Base-T4
100Base-TX
1000Base-LX
→ 10Mbps, baseband, over two twisted-pair cables
→ 100Mbps, baseband, over two twisted-pair cables
→ 100Mbps, baseband, over four twisted-pair cables
→ 1000Mbps, baseband, long wavelength over optical fiber cable
113
Ethernet Physical Layer
● In Ethernet, the middle term always seems to be “Base”.
● Early versions of protocol also allowed for broadband transmissions
(example, 10Broad)
○
Broadband uses analog signals in the form of optical or electromagnetic waves over multiple
transmission frequencies.
● The broadband implementations were not successful in marketplace
● Hence, all current Ethernet implementations use baseband transmissions
○
Baseband uses digital signaling over a single wire, and the transmission takes form of either
electrical pulses or light
114
Twisted-Pair Cable
● It consists of two insulated copper
wires that are twisted together in
a helical form
○
Twisting is necessary because two
parallel wires constitute a finite antenna.
When wires are twisted,
waves from each cancel
out, so the wire radiates
less effectively.
Category 5 UTP cable with 4 twisted pairs
115
Twisted-Pair Cables
●
Category 3 cables are replaced by Category 5 cables
○
○
●
●
Cat.5 has more twists per meter. More twists result in less
crosstalk and better-quality signal over longer distances.
Hence, Cat.5 is more suitable for high-speed computer
communications
Category 5 cabling consists of two insulated wires
gently twisted together. Four such wires are grouped in
a plastic sheath to protect the wires and keep them
together.
Cat.6 and Cat.7 have more stringent specifications to
handle signals with greater bandwidths.
Category 5
cable
116
Coaxial Cable
●
●
●
●
It has better shielding and greater bandwidth than twisted pairs
It can span longer distances at higher speeds
It has good combination of high bandwidth and excellent noise immunity
Two types:
○
○
50-ohm cable: usually for digital transmission.
75-ohm cable: usually for analog transmission and cable television.
117
10Base-T
● It provides Manchester-encoded 10-Mbps bit-serial communication over two
unshielded twisted-pair cables.
● In Manchester encoding, each pulse
is clearly identified by the direction of
the mid-pulse transition rather than by
its sampled level value.
● Manchester encoding introduced some difficult frequency-related problems
that make it unsuitable for use at higher data rates.
118
10Base-T
●
●
It uses two pairs of a four-pair category 3
or 5 cable. One for sending ad one for
receiving
Each pair is configured as a simplex link
○
●
●
Simplex link allows transmission in one
direction only
10Base-T physical layers can support
either half-duplex or full-duplex
operation.
Full Duplex is supported by using one pair
for sending and one pair for Receiving
119
100Mbps-Fast Ethernet
● Ethernet versions with different encoding requirements and a different set of
media-dependent sublayers:
○
○
○
100Base-T4 (1995)
100Base-TX (1995)
100Base-T2 (97). Research, never commercialized
120
100Base-T4
●
●
●
●
●
It allows 10Base-T networks to be
upgraded to 100Mbps operation
without the need of changing the
existing Cat.3 UTP cables to Cat.5
cables
It uses the the same precedent
signal transmission procedure (2
simplex and 2 half-duplex)
Full-duplex is unsupported on
100Base-T4.
Bit rate /pair = 100/3 = 33.3 Mbps.
Baud rate /pair= 33.33 *6/8 = 25M
baud
121
100Base-T4: 8B6T Encoding
●
●
●
8B6T encoding maps each 8-bit
binary byte into a pattern of 6
ternary (three-level: +, 0, -)
symbols known as 6T codegroups
6T gives 36 = 729 values, we need
only 256.
Separate code-groups are used
for IDLE and SOF, EOF=>no bit or
byte stuufing, control code-groups.
Data (Hex)
(Binary)
8B6T Code
00
0000 0000
+-00+-
01
0000 0001
0+-+-0
...
...
...
FE
1111 1110
-+0+00
FF
1111 1111
+0-+00
122
100Base-T4: 8B6T Encoding
●
●
●
8B6T encoding maps each
8-bit binary byte into a
pattern of 6 ternary (threelevel: +, 0, -) symbols known
as 6T code-groups
6T gives 36 = 729 values, we
need only 256.
Separate code-groups are
used for IDLE and SOF,
EOF=>no bit or byte
stuufing, control codegroups.
123
100Base-T4: 8B6T Encoding
●
●
●
During transmission, 6T code-groups are transmitted in a delayed round-robin
sequence over the 3 wires
Each frame is encapsulated with Start-of-Stream and End-of-Stream code-groups.
Receipt of a non-IDLE code-group over the dedicated receive-pair any time before
the collision window
expires indicates
that a collision has
occurred.
124
100Base-X(100Base-TX)
● It supports transmission over either two pairs of Category 5 UTP copper wire or two
●
●
strands of optical fiber.
Full Duplex is Supported by using one pair for sending and one pair for Receiving.
The 100Base-X uses the 4B/5B encoding
○
Each 4-bit data nibble (representing half of a data byte) is mapped into a 5-bit binary codegroup that is transmitted bit-serial over the link
● The 32 5-bit code-groups are divided as (See Table Next page):
○
○
○
○
16 possible values in a 4-bit data nibble
4 control code-groups transmitted as pairs (Start of Stream, End of Stream)
1 special IDLE code-group that is continuously sent during interframe gaps to maintain
continuous synchronization between NICs
11 invalid code-groups. If any is included in the transmission, the frame is treated as invalid
125
100Base-X: 4B/5B
Encoding Table.
126
100Base-X: Stream with Frame Encapsulation
127
100Base-T2
● It is a better alternative for
upgrading networks with
installed Cat.3 cabling
● It uses dual-duplex baseband
transmission method to send
encoded symbols
simultaneously in both
directions on both wire pairs.
● Was developed in 1997.
Research only. Never
comercialized
128
100Base-T2 (Full Duplex Operation)
● Because the signal on each wire pair at the
MDI (Medium Interface) is the sum of the
transmitted signal and the received signal,
each receiver subtracts the transmitted
symbols from the signal received at the MDI
to recover the symbols in the incoming data
stream. The incoming symbol pair is then
decoded, unscrambled, and reconstituted as a
data nibble for transfer to the MAC.
● Was not commercialized. But Ideas were used
in Giga bit Ethernet
129
100Base-T2: PAM5 Encoding
● First, data frame nibbles is scrambled to randomize the bit sequence
● PAM5 (Pulse Amplitude Modulated) maps the two upper bits and the two
lower bits of each nibble into 5-level (+2, +1, 0, -1, -2) PAM symbols.
● The mapped nibbles are transmitted simultaneously over the two wire pairs
(PAM5x5)
● Scrambling of data ensures that the data streams traveling in opposite
directions on the same wire pair are uncoordinated
130
100Base and 10Base Table
Values for bit and Baud rate are per pair
131
1000Base-T
● It scrambles each byte in the MAC frame to randomize the bit sequence
before it is encoded using 4 PAM5 symbols
● The PAM5 symbols are sent at the same time over 4 wire pairs
● 4 of the 5 levels in each PAM5
symbol represent 2 bits in the
data byte
● The 5th level is used for FEC
(Forward Error Correction)
coding
○
This enhances symbol recovery
in the presence of noise and crosstalk
132
1000Base-T
•
1000Base-T supports full-duplex
operations and of course half duplex.
•
Full-duplex: Because the signal on each
wire pair at the MDI (Medium Interface) is
the sum of the transmitted signal and the
received signal, each receiver subtracts the
transmitted symbols from the signal
received at the MDI to recover the symbols
in the incoming data stream. The incoming
symbol pair is then decoded, unscrambled,
and reconstituted as a data nibble for
transfer to the MAC.
133
1000Base-T
Cat 5e STP or FTP
134
1000Base-X
● It supports full-duplex binary transmission over
two strands of optical fiber or two STP cables.
○
It supports as well half-duplex operation
● It uses 8B/10B encoding scheme
○
Each 8-bit data byte is mapped into a 10-bit
code-group for bit-serial transmission
● Each frame is encapsulated at physical layer
● Link synchronization is maintained by sending
a continuous stream of IDLE code-groups during interframe gaps
135
1000Base-X
● It supports full-duplex binary transmission over
two strands of optical fiber or two STP cables.
○
It supports as well half-duplex operation
● It uses 8B/10B encoding scheme
○
Each 8-bit data byte is mapped into a 10-bit
code-group for bit-serial transmission
136
Fiber Optics
● A high-speed transmission medium with clear signal
● The current practical limit of bandwidth of fiber optics is 100 Gbps
○
The achievable (theoretical) bandwidth with fiber technology is in excess of 50,000 Gbps (50
Tbps). This limits are unreachable due to the inability to convert between electrical and optical
signals any faster
● To build higher-capacity links, many channels are simply carried in parallel
over a single fiber.
● Fiber optics are used for long-distance transmission in network backbones
○
As in high-speed LANs and high-speed Internet access such as FttH (Fiber to the Home)
137
Fiber Optics
● An optical transmission has three components:
○
○
○
The light source
■ A pulse of light indicates a 1 bit and the absence of light indicates a 0 bit
The transmission medium
■ It is an ultra thin fiber of glass
The detector
■ It generates an electrical pulse when light falls on it
● By attaching a light source to one end of an optical fiber and a detector to the
other, we have a unidirectional data transmission system that accepts an
electrical signal, converts and transmits it by light pulses, and then reconverts
the output to an electrical signal at the receiving end
138
Fiber Optics
● In this transmission system, the light should not leak and be useless.
○
○
○
○
If a light ray passes from one medium to another, the ray is bent at the boundary.
The amount of refraction depends on the properties of the two media
If the angles of incidence are above a certain critical value, the light is refracted back into the
silica and none will escape into the air (trapped inside the fiber).
The light inside the fiber can propagate for many kilometers with virtually no loss
139
Fiber Optics: Modes
● Multi-mode fiber
○
○
Any light ray incident on the boundary above the
critical angle will be reflected internally
Many different rays can be sent with different angles
● Single-mode fiber
○
○
○
The fiber’s diameter is reduced to a few wavelengths
of light
Hence, the fiber acts like a wave guide and the light
can propagate only in a straight line without bouncing
It is more expensive and used for longer distances
■ data transmission at 100 Gbps for 100 km
without amplification
140
Transmission of Light Through Fiber
● Optical fiber is made of glass that is transparent enough for the light to shine
through
● The attenuation of light through glass is defined as the ratio of input to output
signal power
● The attenuation of light depends on:
○
○
the wavelength of the light
the physical properties of the glass
141
Transmission of Light Through Fiber
0.85 band has
higher
attenuation and
is used for
shorter
distances. But at
this wavelength,
lasers and
electronics are
made of the
same material.
These two bands
have good
attenuation
properties (less
than 5% loss per
kilometer)
The attenuation of light through fiber in the infrared region
142
Fiber Cables
● Fiber cables are almost like coaxial
cables except for the braided outer
conductor.
● The center is the fiber core through
which the light propagates
○
○
In multimode fibers, the core is typically
50 microns in diameter
In single-mode fibers, the core is 8 to 10 microns
● The cladding glass is used to keep all the light in the core
● The plastic jacket is used to protect the cladding
143
Fiber Cables
● Fiber is normally grouped in bundles
protected by an outer sheath
● Fibers are connected in three ways:
○
○
○
using fiber sockets.
■ Connectors lose about 10-20% of the light but
they make it easy to reconfigure
using mechanical splices to attach two connectors
by using a special sleeve
■ about 10% of light loss
by melting two connectors together to form a solid
connection
■ small amount of attenuation occurs
144
Light Sources
● There are two types of light sources:
○
○
LED (Light Emitting Diode)
Semiconductor lasers(ILD: Injection Laser Diode)
● The receiving end
is a photodiode
○
The response time
of photodiode limits
the data rate to
about 100 Gbps
145
Optical Fiber vs. Copper Wire
● Advantages of optical fiber over copper wires
○
○
○
○
○
○
It can handle much higher bandwidths than copper
It has low attenuation, so less repeaters are needed
■ repeaters every 50 km versus every 5 km for copper
It is not affected by electromagnetic interference, power failures and corrosive chemicals
It is thin and lightweight
■ 1000 twisted-pair wires 1 km long weight 8000 kg
■ two fiber optics have more capacity than the 1000 twisted pairs and they weight 100 kg
Lower costs regarding support systems to maintain them and lower installation costs of new
routers
More Secure. No induced EMF.
146
Optical Fiber vs. Copper Wire
● Advantages of copper wire over optical fiber
○
○
○
○
○
Fiber is a less familiar technology requiring skills not all engineers have
Fiber can be damaged easily by being bent too much
Optical transmission is unidirectional
■ Two-way communication requires either two optical fibers or two frequency bands on
one fiber
Fiber interfaces cost more than copper interfaces
More Secure
147
1000-X (SX, LX, CX) Ethernet
148
1000 Base-X Ethernet
Multimode and Single Mode Fiber
149